Problem: Evaluate $~~\int^e_1\dfrac{\ln x}{x^3}\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-3+e^2}{4e^2}$ (Choice B) B $\dfrac{1+e^2}{4e^2}$ (Choice C) C $\dfrac{-1+2e^2}{4e^2}$ (Choice D) D $\dfrac{-3+2e^2}{4e^2}$
Explanation: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv= \dfrac1{x^3}\,dx\,$. Then $~du = \dfrac1xdx~$ and $~v = -\dfrac1{2x^2}\,$. Integration by parts gives $ \int^e_1 \dfrac{\ln x}{x^3}dx =-\dfrac{\ln x}{2x^2}\Bigg]_1^e+\int^e_1\dfrac1{2x^3}dx$ $ ~~=-\dfrac{\ln x}{2x^2}-\dfrac1{4x^2}\Bigg]^e_1=\Bigg(-\dfrac{1}{2e^2}-\dfrac1{4e^2}\Bigg)-\Bigg(0-\dfrac{1}{4}\Bigg)$ $ ~~=\dfrac{-3+e^2}{4e^2}$